Question

1. The nozzle shown in the sketch below is designed to accelerate the flow of CO2 from a velocity close to zero to one that is rather high. The inlet temperature and pressure are 1000K and 614.705 kPa, respectively, and the nozzle exhausts to the atmosphere at 100 kPa. Using the ideal Gas Tables Determine the outlet velocity and temperature when the expansion through the nozzle is considered to be ideal (reversible); a. b. Calculate the temperature and velocity at the nozzle exit when the isentropic efficiency of the nozzle is 87.56385% WATCH YOUR UNITS!!! T1 = 1000 K 1 p, 614.705 kPa P2 100 kPa

Answer 1

Here,

T₁= 1000K, P₁= 614.705kPa And P₂= 100kPa.

From the table, Entropy for T_1, S₀1 = 269.215 kJ/kmol-K

S₀2= S₀1 + R ln (P₂/P1)

= 269.215 + (8.314/44) ln(100/614.705)

= 269.215-0.3429

=268.87 kJ/kmol-K

For the above value of S₀2, value of T₂ from the table= 990 K.

Velocity at outlet V₂= [(h₁-h₂) * 2g_c]

Here, h₁ and h₂ are specific enthalpies at temperature T₁ and T_2.

From the table, h₁= 42,769 kJ/kmol and h₂= 42,226 kJ/kmol.

Thus, V₂ = [(42769-72226)*2(1)]^1/2

= 32.954 m/s.